A straight line is given as 2(x+4) = -2(y-5) = -3(z-6),
(a)determine the vector equation of the straight line. The answer I found is r = -4i +5j +6k + t(3i -3j +2k), is this correct?
(b)find the coordinate of a point Q that lies at the straight line such the |(OQ) ⃗ |= 3. I can't get this one, someone please guide me, thanks.
With t= 1, r= -4i+ 5j+ 6k+ 3i- 3j+ 2k= -i+ 2j+ 8k or the point (-1, 2, 8). 2(x+ 4)= 2(-1+ 4)= 6, -2(y- 5)= -2(2- 5)= 6, and -3(z- 6)= -3(8- 6)= -6. Those are NOT equal so the point (-1, 2, 8) on your line is not on the given line. They are not the same line.
I presume you know that the distance from Q= (x, y, z) to O= (0, 0, 0) is so, setting that equal to 3 and squaring, .(b)find the coordinate of a point Q that lies at the straight line such the |(OQ) ⃗ |= 3. I can't get this one, someone please guide me, thanks.
Since all of 2(x+ 4), -2(y+ 5), and -3(z- 6) are equal, set them all equal to t: 2(x+ 4)= t so x= -4+ (1/2)t. -2(y+ 5)= t so y= -5- (1/2)t. -3(z- 6)= t so z= 6- (1/3)t. Replace x, y, and z in to get a single equation in t.